Given the oft-cited statistic that 66.7% of the American public is overweight or obese, you are interested in comparing college students BMI level to the general population.
Assuming that the general population has a BMI of (μ) = 27.5 using α level of .05, you hypothesize that college students will have a lower BMI than the general population.
State the null and alternative hypothesis and the critical value. Conduct a single-sample t-test based on the following data. Provide an approximate p-value and your conclusion.
Also, what would be the critical value if this was a two-tailed test and what would be your conclusion regarding reject/fail to reject the null hypothesis?
Given this, provide a 95% confidence interval (for the single sample t-test we’ll do the CI around the sample mean, though you can also do it around the difference between the sample mean and population mean).
Further, what shape does the distribution of scores take? What is your conclusion?
BMI
24.00
28.00
22.00
28.00
24.00
21.00
23.00
32.00
25.00
23.00
22.00
21.00
30.00
21.00
22.00
You'll need to calculate the mean and standard deviation. Both of these values will be needed in the following formula:
t = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
Ho: µ = 27.5 -->null hypothesis
Ha: µ < 27.5 -->alternative hypothesis
Use a t-table at 0.05 level of significance for a one-tailed test (alternative hypothesis shows a specific direction) at 14 degrees of freedom (df = n - 1 = 15 - 1 = 14).
Once you have the critical value from the table, compare the test statistic to the critical value after you finish the calculations from the formula. If the test statistic exceeds the critical value from the table, reject the null. If the test statistic does not exceed the critical value from the table, do not reject the null. Do the same for a two-tailed test. Remember to check the t-table for two-tailed to obtain the correct critical values (there will be a plus and minus value since this will be two-tailed). To find the p-value, use the table again. The p-value is the actual level of the test statistic.
To find the 95% confidence interval around the sample mean, use the critical values from the two-tailed test in the appropriate confidence interval formula. Draw your conclusions from the results you find.
I hope these few hints will help get you started.
This helps a LOT. Thank you very much for your time.
To compare college students' BMI levels to the general population, we can conduct a single-sample t-test. First, let's state the null and alternative hypotheses:
Null hypothesis (H0): The average BMI of college students is equal to the general population (μ = 27.5).
Alternative hypothesis (H1): The average BMI of college students is lower than the general population (μ < 27.5).
Next, we need to determine the critical value. Since the alpha (α) level is given as 0.05, this is a one-tailed test, and we will use the t-distribution with n-1 degrees of freedom. The critical value corresponds to the alpha level and degrees of freedom.
Using a t-table or t-distribution calculator, with an alpha level of 0.05 and degrees of freedom (n-1) = 14-1 = 13, we find the critical value to be -1.771.
Now, let's calculate the single-sample t-test using the provided data:
BMI data:
24.00
28.00
22.00
28.00
24.00
21.00
23.00
32.00
25.00
23.00
22.00
21.00
30.00
21.00
22.00
Step 1: Calculate the sample mean.
μ = (24.00 + 28.00 + 22.00 + 28.00 + 24.00 + 21.00 + 23.00 + 32.00 + 25.00 + 23.00 + 22.00 + 21.00 + 30.00 + 21.00 + 22.00) / 15
μ = 25.20
Step 2: Calculate the sample standard deviation.
s = √[((24.00-25.20)^2 + (28.00-25.20)^2 + ... + (22.00-25.20)^2) / (15-1)]
s ≈ 3.033
Step 3: Calculate the t-value.
t = (sample mean - population mean) / (sample standard deviation / √n)
t = (25.20 - 27.5) / ( 3.033 / √15)
t ≈ -1.333
Step 4: Calculate the p-value (assuming a one-tailed test).
This can be done using a t-table or a statistical software. For t ≈ -1.333, with 13 degrees of freedom, the p-value is approximately 0.104.
Comparing the p-value (0.104) to the given significance level (α = 0.05), we can make the conclusion:
Since the p-value (0.104) is greater than the significance level (0.05), we fail to reject the null hypothesis. This means that there is not enough evidence to support the claim that the average BMI of college students is lower than the general population.
If this were a two-tailed test, the critical value would be the same in absolute value but with a different sign (positive and negative). In this case, the critical value would be ±1.771.
To calculate a 95% confidence interval (CI) for the single-sample t-test, we can use the formula:
CI = (sample mean ± (t-value * (sample standard deviation / √n)))
Using the provided values, substituting t-value as 1.771 and sample mean ± (1.771 * (3.033 / √15)), we can calculate the 95% CI. The lower bound and upper bound of the CI are then subtracted and added to the sample mean, respectively.
CI = 25.20 ± (1.771 * (3.033 / √15))
CI ≈ 25.20 ± 1.49
CI ≈ (23.71, 26.69)
The distribution of scores in this case can be assumed to be approximately normal since the sample size (15) is relatively large (assuming n > 30 is a general rule). This assumption allows us to use the t-test for inference.
In conclusion, based on the single-sample t-test, we fail to reject the null hypothesis, indicating no significant difference between the average BMI of college students and the general population.